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Evaluate the following limit : limx→0[3x+3-x-2x⋅tanx]

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प्रश्न

Evaluate the following limit : 

`lim_(x -> 0) [(3^x + 3^-x - 2)/(x*tanx)]`

योग
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उत्तर

`lim_(x -> 0) [(3^x + 3^-x - 2)/(xtanx)]`

= `lim_(x -> 0) (3^x[3^x + 3^-x - 2])/(3^x*xtanx)`

= `lim_(x -> 0) ((3^x)^2 + 1 - 2(3^x))/(3^x*xtanx)`

= `lim_(x -> 0) (3^x - 1)^2/(3^x *xtanx)`

= `lim_(x -> 0) ((3^x - 1)/x)^2/(3^x * (tanx/x))`  ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]

= `(lim_(x -> 0) (3^x - 1)/x)^2/((lim_(x -> 0) 3^x) xx (lim_(x -> 0) tanx/x)`

= `(log3)^2/(3^0*1)    ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`

= (log 3)2

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अध्याय 7: Limits - Exercise 7.6 [पृष्ठ १५४]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Exercise 7.6 | Q II. (1) | पृष्ठ १५४

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